A glass full of water has a bottom of area `20 cm^(2)`, top of area `20 cm`, height `20 cm` and volume half a litre.

a. Find the force exerted by the water on the bottom.
b. Considering the equilibrium of the `v`, after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)`

Answer

A, B, D

Explanation

`Pa=1.0xx10^5N/m^2`,
`=Pw=10^3mg/m^3`
g=10m/s^2`
`v=500m/=500g=0.5kg`
a. Force exerted at the bottom = Force due to cylindrical
water column + atm. Force
`=AxxhxxP_wxxg+P_axxA`
`=A(hrho_wg+P_a)`
`=204N`
b. To find out the resultant force exerted by the sides of the glass, from the free body diagram of water inside the glass,
`P_axxA+mg=Axxhxxrho_2xxg+F_s+P_axxA`
`rarr mg=Axxhxxp_wxg+F_s`
`rarr 0.5xx1=20xx10^-4xx20x10^-2xx10^-3xx10+F_s`
`rarr F_s=5-4=1N` (upward)
This force 1N is provided by the sides of the glass.